An urn contains 3 red and 4 green marbles (7 total). If three marbles are picked at random without replacement, what is the probability that two are green and one is red?
Aptitude
Probability
Difficulty: Medium
Choose an option
-
A3/7
-
B18/35
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C5/14
-
D4/21
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ENone of these
Answer
Correct Answer: 18/35
Explanation
Introduction / Context:Sampling without replacement from colored marbles is modeled by combinations. We count favorable color-compositions and divide by total equally likely 3-combinations from the urn.
Given Data / Assumptions:
- R = 3 (red), G = 4 (green), total N = 7.
- Draw k = 3 without replacement.
- Target composition: 2 green, 1 red.
Concept / Approach:
- Total ways = C(7,3).
- Favorable = C(4,2) * C(3,1).
- Probability = favorable / total.
Step-by-Step Solution:
Total ways = C(7,3) = 35Favorable ways = C(4,2) * C(3,1) = 6 * 3 = 18Probability = 18 / 35Verification / Alternative check:Hypergeometric model: HG(N=7, success-type G=4, draws k=3, want x=2) gives the same ratio 18/35 after multiplying by C(3,1) for the red pick.
Why Other Options Are Wrong:
- 3/7, 5/14, 4/21 result from mixing replacement assumptions or miscounting combinations.
Common Pitfalls:
- Treating order as distinct (it is not; combinations handle orderless draws).
Final Answer:18/35